Below is a problem just like yours with different numbers.
You can use it as a guide to solve yours. Yours has this
solution:
(x,y,z) = (
, 
, 
)
This one has solution (x,y,z) = (15, -49, 27)
x + 2y + 3z = -2
x + y + z = -7
-x + y + 2z = -10
I will assume you already know how to find the inverse
of a matrix, and how to multiply two matrices. If you don't,
post again asking how.
First we form three matrices, A, X, and B.
1. Matrix A is the 3x3 coefficient matrix A, which consists
of just the three columns of x, y, and z coefficients. in
that order, but does not contain the column of constants.
A = 
.
2. Matrix X is the 3x1 matrix of variables X = 
3. Matrix B is the 3x1 matrix, whose only column is the
column of constants: B = 
Next we form the matrix equation:
AX = B
or
To solve the equation
AX = B
we left-multiply both sides by A-1, the inverse of A.
A-1(AX) = A-1B
Then since the associative principle holds for matrix multiplication,
(even though the commutative principle DOES NOT!!!), we can move
the parentheses on the left around the first two matrix factors:
(A-1A)X = A-1B
Now since A-1A = I, where I is the identity matrix, the
above becomes:
IX = A-1B
and by the identity property:
X = A-1B
Performing these operations with the actual matrices we have
the equation AX = B
Next we find the inverse of A, which is written A-1.
A-1 = 
Then we indicate the left multiplication of both sides by
to get the equation A-1(A*X) = A-1B:
Next we use the associative principle to move the parentheses so that
they are around the first two factors to get the equation
(A-1A)X=A-1B:
to get the equation (A-1A)X = A-1B:
When we perform the matrix multiplication we get:
The matrix on the left is the identity matrix
Then when we multiply the identity matrix I by the column matrix of
variables, we just get the matrix of variables, or the
equation X = A-1B
or x=15, y=-49, z=27
Edwin
